Question

A parallel plate capacitor is connected to a battery and is in a steady state. Keeping the battery connected, if the area of the plates is increased then

• A. The potential difference across the plates will remain the same
• B. The magnitude of charge on each of the plates will remain the same
• C. The capacitance of the capacitor will reduce
• D. Electric field between the plates will decrease

Answer 1

The correct option is A The potential difference across the plates will remain the same

With battery connected, the potential difference across the plates of the capacitor will remain equal to the emf of the battery.

So, the potential difference across the plates will remain same.
Therefore, the option (a) is correct.

Capacitance of a parallel plate capacitor is given by $$C=\frac{{\epsilon }_{0}A}{d}$$ As the area of the plates increases, $C$ also increases.
So, option (c) is incorrect.

Magnitude of charge on the plates is given by $$Q=CV$$ With constant $V$ as $C$ increases, $Q$ will also increase.
Hence, option (b) is incorrect.

Now, electric field is given by $$E=\frac{V}{d}$$As $V$ and $d$ are constant, so $E$ will also remain constant. Hence, option (d) is also incorrect.

Key concept: Voltage across PPC keeping battery connected with the change in plate area .